Codeforces 828C String Reconstruction【思维+并查集】

C. String Reconstruction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan had string s consisting of small English letters. However, his friend Julia decided to make fun of him and hid the strings. Ivan preferred making a new string to finding the old one.

Ivan knows some information about the string s. Namely, he remembers, that stringt_i occurs in strings at least k_i times or more, he also remembers exactlyk_i positions where the stringt_i occurs in strings: these positions are x_i, 1, x_i, 2, ..., x_i, ki. He remembersn such strings t_i.

You are to reconstruct lexicographically minimal strings such that it fits all the information Ivan remembers. Stringst_i and strings consist of small English letters only.

Input

The first line contains single integer n (1 ≤ n ≤ 10⁵) — the number of strings Ivan remembers.

The next n lines contain information about the strings. Thei-th of these lines contains non-empty stringt_i, then positive integerk_i, which equal to the number of times the stringt_i occurs in strings, and then k_i distinct positive integersx_i, 1, x_i, 2, ..., x_i, ki in increasing order — positions, in which occurrences of the string t_i in the strings start. It is guaranteed that the sum of lengths of stringst_i doesn't exceed10⁶, 1 ≤ x_i, j ≤ 10⁶,1 ≤ k_i ≤ 10⁶, and the sum of allk_i doesn't exceed10⁶. The strings t_i can coincide.

It is guaranteed that the input data is not self-contradictory, and thus at least one answeralways exists.

Output

Print lexicographically minimal string that fits all the information Ivan remembers.

Examples

Input

3a 4 1 3 5 7ab 2 1 5ca 1 4

Output

abacaba

Input

1a 1 3

Output

aaa

Input

3ab 1 1aba 1 3ab 2 3 5

Output

ababab

题目大意：

现在让你构造出来一个最小字典序的字符串，使得其满足n个条件。

这n个条件中每个条件有一个连续的子字符串。

然后有Ki个位子，表示这些字符串出现的位子。

保证有解，n个条件互相没有矛盾。

思路：

问题其实就是一个涂色问题。每次在Ki个点处作为起点，然后向后涂这个子字符串，如果已经涂过的位子不用再涂，否则涂上这个字母。

那么暴力去涂色肯定是要超时的。

我们维护一个并查集，使得find（x）查找到的位子是位子x之后，第一个没有涂色过的位子。

那么我们在暴力涂色的过程中，可以用并查集优化寻找速度。

很古老的套路题了= =

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int f[3000000];char ans[3000000];char a[3000000];int find(int a){    int r=a;    while(f[r]!=r)    r=f[r];    int i=a;    int j;    while(i!=r)    {        j=f[i];        f[i]=r;        i=j;    }    return r;}void merge(int a,int b){    int A,B;    A=find(a);    B=find(b);    if(A>B)    {        f[B]=A;    }    else f[A]=B;}int main(){    int n;    while(~scanf("%d",&n))    {        memset(ans,' ',sizeof(ans));        int end=0;        for(int i=1;i<=3000000-80;i++)f[i]=i;        for(int i=1;i<=n;i++)        {            scanf("%s",a);            int lenn=strlen(a);            int ki;            scanf("%d",&ki);            while(ki--)            {                int x;                scanf("%d",&x);                int tmpx=x;                end=max(end,lenn+x-1);                while(x<=tmpx+lenn-1)                {                    ans[x]=a[x-tmpx+1-1];                    x=find(x);                    if(x<=tmpx+lenn-1)merge(x,x+1);                }            }        }        for(int i=1;i<=end;i++)        {            if(ans[i]==' ')ans[i]='a';            printf("%c",ans[i]);        }        printf("\n");    }}